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#### Ayush starts walking from his house to the office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2,4), bank at (5, 8), school at (13, 14) and office at(13, 26) and coordinates are in km.

The given point are (2, 4), (5, 8), (13, 14), (13, 26)
Distance between house and bank

$= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$= \sqrt{\left ( 5-2 \right )^{2}+\left ( 8-4 \right )^{2}}= \sqrt{9+16}= 5$
Distance between bank and school
$= \sqrt{\left ( 13-5 \right )^{2}+\left ( 14-8 \right )^{2}}= \sqrt{64+36}= \sqrt{100}= 10$
Distance between school and office
$= \sqrt{\left ( 13-13 \right )^{2}+\left ( 26-14 \right )^{2}}= \sqrt{\left ( 12 \right )^{2}}= 12$
Distance between office and house
$= \sqrt{\left ( 13-2 \right )^{2}+\left ( 26-14 \right )^{2}}$
$= \sqrt{121+484}$
$= \sqrt{605}$
$= 24\cdot 59$

Total distance covered from house to bank, bank to school, school to office = 5 + 10 + 12 = 27
Extra distance covered = 27 – 24.59 = 2.41 km.

#### Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

Solution.         Points are A(3, 5), B(7, 9), C(11, 5), D(7, 1)
$Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$AB= \sqrt{\left ( 7-3 \right )^{2}+\left ( 9-5 \right )^{2}}$
$AB= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}$
$Length\, of\, BC= \sqrt{\left ( 11-7 \right )^{2}+\left ( 5-9 \right )^{2}}$
$BC= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, CD= \sqrt{\left ( 7-11 \right )^{2}+\left ( 1-5 \right )^{2}}$
$CD= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, AD= \sqrt{\left ( 3-7 \right )^{2}+\left ( 5-1 \right )^{2}}$
$AD= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, AC= \sqrt{\left ( 11-3 \right )^{2}+\left ( 5-5 \right )^{2}}$
$AC= \sqrt{\left ( 8 \right )}$
= 8
$Length\, of\, BD= \sqrt{\left ( 7-7 \right )^{2}+\left ( 1-9 \right )^{2}}$
$BD= \sqrt{64}$
= 8
AB = BC = AD, AC = BD
Hence, ABCD is square
The diagonals cut each other at mid-point, which is the equidistance from all four corners of square.
$Mid-point \, of \, AC = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$
(x1, y1) =(3, 5)             (x2, y2) = (11, 5)
$AC= \left ( \frac{3+11}{2},\frac{5+5}{2} \right )$
$AC= \left ( 7,5 \right )$
This should be the position of Jaspal.

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#### If the points A (1, –2), B (2, 3) C (a, 2) and D (– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Solution

We know that diagonals bisect each other
Hence, mid-point of AC = mid-point of BD
$\left ( \frac{1+a}{2},\frac{-2+2}{2} \right )= \left ( \frac{2-4}{2},\frac{3-3}{2} \right )$
$\left ( \frac{1+a}{2},0 \right )= \left ( -1,0 \right )$
$\frac{1+a}{2}= -1$
1 + a = –2
a = –3
C(–3, 2)
$Area \, of \, \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 1\left ( 3-2 \right ) +2\left ( 2+2 \right )+\left ( -3 \right )\left ( -2-3 \right )\right ]$
$= \frac{1}{2}\left [ 1+2\left ( 4 \right )+15\right ]$
$= \frac{1}{2}\left [ 24 \right ]= 12sq\cdot units$
Area of parallelogram = 2 × Area of $\triangle$ABC
Area of parallelogram = 2 × 12 = 24sq.units
$Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$= \sqrt{\left ( 2-1 \right )^{2}+\left ( 3+2 \right )^{2}}$
$AB=\sqrt{1+25}= \sqrt{26}units$
Area of parallelogram = Base × height
$\frac{24}{Base}= Height$
$Height= \frac{24}{AB}$
$Height= \frac{24}{\sqrt{26}}\times\frac{\sqrt{26}}{\sqrt{26}}$
$\frac{24\sqrt{26}}{13}units$

#### (i) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.The median from A meets BC at D. Find the coordinates of the point D. (ii) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.Find the coordinates of the point P on AD such that AP : PD = 2 : 1 (iii) The points A(x1, y1)B(x2, y2)and C(x3, y3)are the vertices of $\triangle$ABC.Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1 (iv) The points A(x1, y1),B(x2, y2) and C(x3, y3) are the vertices of $\triangle$ABC.What are the coordinates of the centroid of the triangle ABC?

(i) Solution

D is the mid-point of BC.
$Mid-point \,formula = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$

Coordinates of $D\left ( x,y \right )= \left ( \frac{x_{2}+x_{3}}{2} ,\frac{y_{2}+y_{3}}{2}\right )$ (By midpoint formula)
(ii) Solution

$Section \, formula= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$D= \left ( \frac{x_{2}+x_{3}}{2},\frac{y_{2}+y_{3}}{2} \right )$
(By Midpoint formula)
$P= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$P= \left ( \frac{2\times \frac{\left ( x_{2}+x_{3} \right )}{2}+1\times x_{1}}{2+1}, \frac{2\times \frac{\left ( y_{2}+y_{3} \right )}{2}+1\times x_{1}}{2+1} \right )$
$P= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iii) Solution

E is mid-point of AC
$E = \left ( \frac{x_{1}+x_{3}}{2} ,\frac{y_{1}+y_{3}}{2}\right )$
Q divides BF at 2 : 1
$Q= \left ( \frac{2\times \frac{\left ( x_{1}+x_{3} \right )}{2}+1\times x_{2}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{3} \right )}{2}+1\times y_{2}}{2+1} \right )$
$Q= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$

R divides CF at 2 : 1
$R= \left ( \frac{2\times \frac{\left ( x_{1}+x_{2} \right )}{2}+1\times x_{3}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{2} \right )}{2}+1\times y_{3}}{2+1} \right )$
$R= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iv) solution

$co-ordinate\, of\,centroid= \left ( \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}, \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}\right )$
Centroid:
The centroid is the centre point of the triangle which is the intersection of the medians of a triangle.
$\triangle ABC\, coordinates\, of\, centroid= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$

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#### A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $\triangle$ADE.

Solution.
$Distance \, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

The given points A(6, 1), B(8, 2) and C(9, 4) let D(x, y)

As the diagonal of a parallelogram bisect each other.
Here mid-point of AC = mid-point of BD
$\left ( \frac{6+9}{2},\frac{1+4}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )$
$\left ( \frac{15}{2},\frac{5}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )$
$\frac{15}{2}= \frac{8+x}{2}$       $\frac{2+y}{2}= \frac{5}{2}$

x = 7         y = 3
D (7, 3)
E is the mid-point of CD
Let E(x0, y0)
$\left ( x_{0},y_{0} \right )= \left ( \frac{7+9}{2}, \frac{3+4}{2} \right )$
$\left ( x_{0},y_{0} \right )= \left ( \frac{16}{2},\frac{2}{7} \right )$
$E= \left ( 8,\frac{7}{2} \right )$
$Area \, of \, \triangle ADE = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 6\left ( \frac{7}{2}-3 \right ) +8\left ( 3-1 \right )+7\left ( 1-\frac{7}{2} \right )\right ]$
$= \frac{1}{2}\left [ 6\times \frac{1}{2}+8\times 2+7\times \frac{-5}{2}\right ]$
$= \frac{1}{2}\left [ 3+16-\frac{35}{2}\right ]$
$= \frac{1}{2}\left [ \frac{6+32-35}{2}\right ]$
$Area\, of\, \triangle ADE= \frac{1}{2}\times \frac{3}{2}= \frac{3}{4}\, sq\cdot units$

#### If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Solution

In equilateral triangle AB = BC = AC
$AC= \sqrt{\left ( x+4 \right )^{2}+\left ( y-3 \right )^{2}}$
$AC= \sqrt{x^{2}+16+8x+y^{2}+9-6y}$
$\left ( because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\; \;\left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
$BC= \sqrt{\left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}}$
$BC= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$      $\left ( because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
AC = BC
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$

Squaring both side
8x + 8x = 0
16x = 0
x = 0
C = (0, y)
$Length\, of\, AB= \sqrt{\left ( 4+4 \right )^{2}+\left ( 3-3 \right )^{2}}$
$AB= \sqrt{\left ( 8 \right )^{2}}= 8$
AC = AB
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= 8$

Put x = 0, squaring both side
0 + 16 + 0 + y2 + 9 – 6y = 64
y2 – 6y + 25 – 64 = 0
y2 – 6y – 39 = 0
$y= \frac{6\pm \sqrt{36+156}}{2}$
$y=\frac{6-8\sqrt{3}}{2}$       (For origin in the interior we take the only term with negative sign)
$y=3-4\sqrt{3}$

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#### ABCD is a quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.

And AB = AD, CB = CD

To prove : AC is $\perp^{r}$ bisector at BD

In $\triangle$ABC and $\triangle$ADC

BC = CD                     (Given)

AC = AC                     (Common)

$\triangle$ABC $\cong$ $\triangle$ADC           (by SSS congruency)

Then by CPCT,

ÐBAC = ÐDAC          …(i)

Now in $\triangle$ABO and $\triangle$ADO

AO = AC                     (common)

$\angle$BAO = $\angle$DAO (from i)

$\triangle$ABO $\cong$ $\triangle$DAO          (by SAS congruency)

Then by CPCT,

$\therefore$ $\angle$AOB = $\angle$DOA

$\because$$\angle$BOD = 180°

$\angle$AOB + $\angle$AOD = 180°                     ($\because$$\angle$BOD = $\angle$AOB + $\angle$AOD)

$\angle$AOB + $\angle$AOB = 180°                      ($\because$$\triangle$AOB $\cong$ $\triangle$AOD)

$\angle$AOB = 90°                                       ($\angle$AOB = $\angle$AOD)

Hence, proved.

#### Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle

Given: ABC is triangle and AC is longest side

To Prove: $\angle ABC > \frac{2}{3}\times 90^{\circ}$

Proof: AC is longest side

$\Rightarrow$$\angle$ B > $\angle$C                                                    (angle opposite to longest side is greaten)

$\angle$B > $\angle$A

$\angle$B + $\angle$B > $\angle$C + $\angle$A

2$\angle$B > $\angle$C +$\angle$A

2$\angle$B + $\angle$B > $\angle$C + $\angle$A + $\angle$B             (adding $\angle$B to LHS and RHS)

3$\angle$B >$\angle$C + $\angle$A + $\angle$B

3$\angle$B > 180°                                                     (Sum of all interior angles in triangle is 180°)

$\angle$B > 60°

$\Rightarrow \angle B > \frac{2}{3}\times 90^{\circ}$

Hence proved.

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#### AB and CD are the smallest and largest sides of a quadrilateral ABCD. Out of $\angle$B and $\angle$D decide which is greater

AB and CD are smallest and greatest side of ABCD.

Join BD

In $\triangle$BCD

DC < BC                                 ($\because$ CD is Greatest side)

$\angle$CBD > $\angle$BDC          (Angle opposite to greater side is greater)                …(i)

In $\triangle$ABD

AD > AB                                 (AB is the smallest side)

$\angle$ABD > $\angle$ADB         (Angle opposite to greater side is greater)                …(ii)

$\angle$CBD + $\angle$ABD > $\angle$BDC + $\angle$ADB

$\angle$ABC >$\angle$ADC

Hence $\angle$B is greater.

#### ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that   AC + AD = BC.

Given, ABC is right angle triangle

AB = AC, CD is bisector at AB

To prove: AC + AD = BC

Proof: $\because$AB = AC (Given)

By Pythagoras theorem

BC2 =AB2 + AC2 = AC2 + AC2 = AC2 + AC2              ($\because$ AB = AC)

$BC =\sqrt{2}AC$

By angle bisector theorem:

$\frac{AD}{BD}=\frac{AC}{BC}=\frac{AC}{\sqrt{2}AC}=\frac{1}{\sqrt{2}}\left ( From \;above \right )$

Let AB = a, AD = b

$\frac{b}{a-b}=\frac{1}{\sqrt{2}}\\ \Rightarrow b=\frac{a}{1+\sqrt{2}}\\ \Rightarrow b=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}$

AC + AD = BC                        ($\because$ AB = AC)

$\Rightarrow \frac{b}{a-b}-\frac{a}{a\sqrt{2}}=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}$

We know, AC = a, AD = b, BC = $a\sqrt{2}$

=> AC + AD = BC

Hence proved

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